Question: Let $a(x)=-6x^3+19x^2+8x+12$, and $b(x)=3x^2+x+1$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Explanation: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's use long division with polynomials in order to find the quotient, $q(x)$ and remainder, $r(x)$ of $\ \dfrac{a(x)}{b(x)}=\dfrac{-6x^3+19x^2+8x+12}{3x^2+x+1}$ : We divide ${3x^2}$ into ${-6x^3}$ to get ${-2x}$ : $ \hphantom{1567|144474} {-2x}\\ {{{3x^2}+x+1}}|\overline{{-6x^3}+19x^2+8x+12}\\ \hphantom{37...8.........|}\llap{-}\underline{(-6x^3-2x^2 -2x)}\\ \hphantom{37|3....998...........}+21x^2+10x \\ $ [What did we do here?] Next, we divide ${3x^2}$ into ${21x^2}$ to get ${+7}$ : $ \hphantom{1567|166664} {-2x \ \ {+ \ \ 7}}\\ {{{3x^2}+x+1}}|\overline{-6x^3+19x^2+8x+12}\\ \hphantom{37...8.........|}\llap{-}\underline{(-6x^3-2x^2 -2x)}\\ \hphantom{37|3....998...........}{+21x^2}+10x+12 \\ \hphantom{37.......888.............|}\llap{-}\underline{(21x^2+7x\ +\ 7)}\\ \hphantom{37|3.............777777.77......}{+3x+5}\\ $ [What did we do here?] The process stops here because $3x^2+x+1$ is a polynomial of the second degree and $3x+5$ is a polynomial of the first degree. So it follows that ${r(x)}={3x+5}$, ${q(x)}={-2x+7}$, and $ \dfrac{-6x^3+19x^2+8x+12}{3x^2+x+1}={-2x+7}+\dfrac{{3x+5}}{3x^2+x+1}$ To conclude, $q(x)=-2x+7$ $r(x)=3x+5$